a,b nguyên dương tm: $\dfrac{ab+1}{a+b} <3/2$
Tìm GTLN bthức:p= $\dfrac{a^3b^3+1}{a^3+b^3} $
1 bài thú vị
Bắt đầu bởi Curi Gem, 01-05-2010 - 14:01
#1
Đã gửi 01-05-2010 - 14:01
4+???=5????
#2
Đã gửi 22-10-2010 - 17:34
ta chung minh hai so co it nhat mot so nho hon hoac = 2. that vay gia su nguoc lai a,b 3 . Ko mat tinh tong quat ta gia su 3 a b.
ta co: $\dfrac{ab+1}{b+a}$ $\dfrac{3b+1}{a+b}$ $\dfrac{3b+1}{2b}=\ frac{3}{2} $
(may thuan voi gia thiet.
gia su a 2 a {1;2}
xet a=1. khi do $P=\dfrac{b^3+1}{b^3+1}=1$
xet a=2.khi do $P=\dfrac{8b^3+1}{b^3+8}=1$
tu dieu kien $ \dfrac{ab+1}{a+b} = \dfrac{2b+1}{b+2} < \dfrac{3}{2}$ $b < 4$
vay b {1;2;3}
voi b=1 ta co $P= \dfrac{9}{9}=1$
voi b=2ta co $P= \dfrac{8.8+1}{8+8} = \dfrac{65}{16}$
voi b=3 ta co $P= \dfrac{27.8+1}{27+8} = \dfrac{217}{35}$ dat duoc khi a=2,b=3 hoac a=3,b=2
ta co: $\dfrac{ab+1}{b+a}$ $\dfrac{3b+1}{a+b}$ $\dfrac{3b+1}{2b}=\ frac{3}{2} $
(may thuan voi gia thiet.
gia su a 2 a {1;2}
xet a=1. khi do $P=\dfrac{b^3+1}{b^3+1}=1$
xet a=2.khi do $P=\dfrac{8b^3+1}{b^3+8}=1$
tu dieu kien $ \dfrac{ab+1}{a+b} = \dfrac{2b+1}{b+2} < \dfrac{3}{2}$ $b < 4$
vay b {1;2;3}
voi b=1 ta co $P= \dfrac{9}{9}=1$
voi b=2ta co $P= \dfrac{8.8+1}{8+8} = \dfrac{65}{16}$
voi b=3 ta co $P= \dfrac{27.8+1}{27+8} = \dfrac{217}{35}$ dat duoc khi a=2,b=3 hoac a=3,b=2
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